Compute conditional probability from two-way tables; apply the addition rule for the probability of A or B; apply the multiplication rule for A and B; and test for independence of two events.

What this topic is asking

The Regents Algebra II exam (the Conditional Probability and the Rules of Probability, S-CP, cluster) wants you to compute conditional probability (often from a two-way table), apply the addition rule for "A or B", apply the multiplication rule for "A and B", and test whether two events are independent. These probability rules and the two-way table are reliable sources of credits.

Conditional probability

Conditional probability answers "given that $B$ happened, how likely is $A$?" by restricting the sample space to outcomes where $B$ occurs.

In a two-way table this is intuitive: $P(A \mid B)$ uses the row or column for $B$ as the new total. If 60 students play a sport and 24 of those also play an instrument, then $P(\text{instrument} \mid \text{sport}) = \frac{24}{60} = 0.4$. The key is using the conditioned group (the 60 sport players) as the denominator, not the whole population.

The addition and multiplication rules

The addition rule subtracts the overlap $P(A \text{ and } B)$ so outcomes in both events are counted once. (For mutually exclusive events, which cannot both happen, the overlap is 0, so it reduces to $P(A) + P(B)$.) The multiplication rule chains the probability of $A$ with the conditional probability of $B$ given $A$.

Independence

Two events are independent when the occurrence of one does not change the probability of the other.

Reading it for credit

A clarifying point that prevents the most common error is the denominator in a conditional probability: $P(A \mid B)$ divides by $P(B)$, so in a table you use the total of group $B$, not the grand total. A second key habit is to subtract the overlap in the addition rule; forgetting it overstates "A or B" by double-counting the shared outcomes. For independence, the Regents accepts either test, comparing $P(A \mid B)$ with $P(A)$, or checking whether $P(A \text{ and } B) = P(A) \cdot P(B)$, but you must show the comparison explicitly, not just assert independence. Stating the conclusion with the supporting equality is what earns the credit.

Try this

Q1. $P(A) = 0.3$, $P(B) = 0.5$, and the events are mutually exclusive. Find $P(A \text{ or } B)$. [1 credit]

• Cue. No overlap, so $0.3 + 0.5 = 0.8$.

Q2. Of 40 people, 10 like both tea and coffee, and 25 like coffee. Find $P(\text{tea} \mid \text{coffee})$. [2 credits]

• Cue. Condition on the 25 coffee drinkers: $\frac{10}{25} = 0.4$.